HDU：A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 204863    Accepted Submission(s): 39378

Problem Description

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

 

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.

 

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.

 

Sample Input

21 2112233445566778899 998877665544332211

 

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110
#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>using namespace std;char a[1001], b[1001], c[1001];int y=0;int main(){    char s1[1001], s2[1001];    int t, cas, i, j, len1, len2;    scanf("%d", &t);    for(cas=1; cas<=t; cas++)    {        memset(a, 0, sizeof(a));        memset(b, 0, sizeof(b));        memset(c, 0, sizeof(c));        cin>>s1>>s2;        len1 = strlen(s1);        len2 = strlen(s2);        for(i=0; i<len1; i++)            a[len1-i-1] = s1[i]-'0';        for(i=0; i<len2; i++)            b[len2-i-1] = s2[i]-'0';        cout<<"Case "<<cas<<":\n";        cout<<s1<<" + "<<s2<<" = ";        if(len1<len2)            len1 = len2;        y=0;        for(i=0;i<len1;i++)            {                y+=a[i]+b[i];                c[i]=y%10+'0';                y=y/10;            }        if(y==1)            cout<<y;        for(i=len1-1; i>=0; i--)            cout<<c[i];        cout<<endl;        if(cas<t)            cout<<endl;    }    return 0;}

#include<iostream>#include<cstdio>#include<string.h>#define N 1005int main(){    int i, j, k, m, l1, l2, t;    char s1[N], s2[N];    int a[N], b[N];    scanf("%d", &t);    for(m=1; m<=t; m++)    {        memset(s1, 0, sizeof(s1));        memset(s2, 0, sizeof(s2));        memset(a, 0, sizeof(a));        memset(b, 0, sizeof(b));        scanf("%s%s",&s1, &s2);        l1=strlen(s1);        l2=strlen(s2);        if(l1>l2)            k=l1;        else            k=l2;        for(i=k,j=l1-1; j>=0; i--,j--)            a[i]=s1[j]-'0';        for(i=k,j=l2-1; j>=0; i--,j--)            b[i]=s2[j]-'0';        for(i=k; i>0; i--)        {            a[i]+=b[i];            if(a[i]>=10)            {                a[i]-=10;                a[i-1]++;            }        }        printf("Case %d:\n%s + %s = ",m,s1,s2);        if(a[0]!=0)        {            for(i=0; i<=k; i++)            {                printf("%d",a[i]);            }        }        else        {            for(i=1; i<=k; i++)                printf("%d",a[i]);        }        if(m<=t-1)            printf("\n\n");        else            printf("\n");    }    return 0;}

JAVA:第一次用JAVA写程序啊。。

import java.math.BigInteger;import java.util.Scanner;public class Main{    public static void main(String[] args)     {        Scanner cin = new Scanner(System.in);        BigInteger a,b,c;        int t=cin.nextInt();        for(int i=1;i<=t;++i)        {            a=cin.nextBigInteger();            b=cin.nextBigInteger();            c=a.add(b);            System.out.println("Case "+i+":");            System.out.println(a + " + " + b + " = " +c);            if(i<t) System.out.println();        }    }}